a1= a+2
a(n+1)=(an+a)/租州√腊型没2
if a=0
a1= 2
a(n+1)=an/√轮纳2
a(n+1)/an = 2^(-1/2)
an/a1 = 2^[-(n-1)/2]
an = 2^[(3-n)/2]
已知数列an满足a1=a+2,an+1=根号2分之an+a，若a=0，求通项公式
a1= a+2
a(n+1)=(an+a)/租州√腊型没2
if a=0
a1= 2
a(n+1)=an/√轮纳2
a(n+1)/an = 2^(-1/2)
an/a1 = 2^[-(n-1)/2]
an = 2^[(3-n)/2]